Question

A stock solution of analyte is made by dissolving 34.83 mg of copper (II) acetate
hexahydrate (fw = 289.73 g/mol) in 25.000 mL of water. A second stock solution of internal
standard is made by dissolving 28.43 mg of germanium (I) acetate (fw = 190.74 g/mol) into
25.000 mL of water. These solutions are used to make a series of standards for flame atomic
absorption analysis calibration. The standard solutions (each 10.00 mL total volume) should
have the following concentrations of copper: 10.00; 25.00; 50.00; 100.0; and 200.0 μM. Each
calibration solution should also contain 50.00 μM of germanium. Based on this information,
complete the following:

Stock solution concentration (analyte): ______________________

Stock solution concentration (internal standard): ______________________

Concentration Cu (μM)	Volume of analyte stock (mL)	Volume of internal standard stock (mL)	Volume of diluent (mL)	Total Volume (mL)
10.00				10.000
25.00				10.000
50.00				10.000
100.0				10.000
200.0				10.000

Answer 1

I. Stock Solution prepared via Copper acetate hexahydrate (289.73g/mol) in 25mL of water

Molarity of Copper acetate solution M1= W*1000/molar mass *Volume

                                                                  = 34.83*10-3*1000/289.73*25

                                                                  = 0.0048 M ~0.005M

II. Internal standard stock solution is prepared via germanium acetate (190.74 g/mol) in 25mL of water

Molarity of Germanium acetate solution M2= W*1000/molar mass *Volume

                                                                  = 28.43*10-3*1000/190.74*25

                                                                  = 0.0059 M ~0.006 M

Dilution:

A) 50 µM germanium acetate as internal standard has to be add in all solutions therefore,

0.0059*X(vol)=50*10-6*10

X=5*10-4/0.0059= 0.0847 mL or 84.7µL internal standard stock has to be added in each tube.

B) Copper acetate stock’s calculation for dilution

M1V1=M2V2 (M1- Copper acetate stock molarity, V1- required amount for specific concentration i.e. M2 and V2- 10 mL)

1) 0.0048* V1=10µM*10

V1=100*10-6/0.0048 = 0.0208 mL or 20.8 µL

2) 0.0048* V1=25µM*10

V1=250*10-6/0.0048 = 0.0520 mL or 52 µL

3) 0.0048* V1=50µM*10

V1=500*10-6/0.0048 = 0.1041 mL or 104 µL

4) 0.0048* V1=100µM*10

V1=1000*10-6/0.0048 = 0.2083 mL or 208.3 µL

5) 0.0048* V1=200µM*10

V1=2000*10-6/0.0048 = 0.5833 mL or 583.3 µL

S.No.	Cu Conc (µM)	Volume of analyte (mL)	Volume of internal standard stock (mL)	Volume of diluent (mL)	Total volume
1	10	0.0208	0.0847	9.8945	10.000 mL
2	25	0.0520	0.0847	9.8633	10.000 mL
3	50	0.1041	0.0847	9.8112	10.000 mL
4	100	0.2083	0.0847	9.707	10.000 mL
5	200	0.5833	0.0847	9.332	10.000 mL